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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following differential equation:
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
✓

Answer

We have,

$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\ (\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{(1+\text{y}^2)}{(\text{x}-\text{e}^{\tan^{-1}\text{y}})}$

$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{x}-\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$

$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\ \dots(1)$

Clearly, it is a linear differential equation of the form

$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$

where

$\text{P}=\frac{1}{1+\text{y}^2}$

$\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$

$\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$

$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$

$=\text{e}^{{\tan^{-1}\text{y}}}$

Multiplying both sides of (1) by $​​​​\text{e}^{\tan^{-1}\text{y}},$ we get

$\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$

$\Rightarrow\ \text{e}^{\tan^{-1}\text{y}}\frac{\text{dx}}{\text{dy}}+\frac{\text{x}\text{e}^{\tan^{-1}\text{x}}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$

Integrating both sides with respect to y, we get

$\text{x}\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy + C}$

$\Rightarrow\ \text{x}\text{e}^{\tan^{-1}\text{y}}=\text{I + C}\ \dots(2)$

Here,

$\text{I}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$

Putting $\tan^{-1}\text{y = t},$ we get

$\frac{1}{1+\text{y}^2}\text{dy = dt}$

$\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$

$=\frac{\text{e}^{2\text{t}}}{2}$

$=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}$

Putting the value of I in (2), we get

$\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}+\text{C}$

$\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+2\text{C}$

$\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ (where K = 2C)

Hence, $2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ is the required solution.

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